NCERT Solutions for Class 12 Chemistry Chapter 1 – Solutions (English Medium)
📘 Get NCERT Solutions for Class 12 Chemistry Chapter 1 – Solutions with accurate answers and step-by-step explanations in a clean, exam-friendly format. This chapter covers types of solutions (solid, liquid, gaseous), concentration terms (mass %, mole fraction, molarity, molality), solubility, Raoult’s law, colligative properties (ΔTb, ΔTf, osmotic pressure), van’t Hoff factor, and abnormal molar mass. Perfect for CBSE board exams and NEET/JEE practice. ✅
🔎 How to search questions?
👉 Use the Search Box below to quickly find any question. Type the Q. number (like Q.1, Q.12, Q.24) or a keyword (like Molarity, Molality, Raoult’s law, Osmotic pressure, Colligative properties, van’t Hoff factor). ✅
- Example: Type Q.10 or Raoult’s law or Molality
- Tip: 2–3 keywords usually find the exact question fast.
- Mobile: You can also use Find in page (Chrome: Menu → Find in page).
Q.1: Define the true solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer
✅
True solution
is a
homogeneous mixture
of
two or more chemically non-reacting substances
(solute +
solvent).👉 In a true solution, solute particles are extremely small (generally < 1 nm) and are uniformly distributed , so the mixture looks the same throughout.
✅ Types of solutions formed: 9 types (based on the physical state of solute and solvent).
Explanation (Step by Step)
🤔 A solution can be classified by what is dissolved
(solute)
and in what it is dissolved
(solvent).So we get 3 solvent states (Gas, Liquid, Solid) × 3 solute states (Gas, Liquid, Solid) = 9 types .
A) Gaseous Solutions (Solvent = Gas)
1) Gas in Gas
👉 Example: Air (O2 + N2 + CO2 etc.)
👉 Example: Air (O2 + N2 + CO2 etc.)
2) Liquid in Gas
👉 Example: Water vapour in air (humidity)
👉 Example: Water vapour in air (humidity)
3) Solid in Gas
👉 Example: Smoke/dust in air, or camphor vapours in N2
👉 Example: Smoke/dust in air, or camphor vapours in N2
B) Liquid Solutions (Solvent = Liquid)
4) Gas in Liquid
👉 Example: CO2 dissolved in water (aerated drinks), O2 dissolved in water (needed by aquatic life)
👉 Example: CO2 dissolved in water (aerated drinks), O2 dissolved in water (needed by aquatic life)
5) Liquid in Liquid
👉 Example: Ethanol in water (alcohol + water mixture)
👉 Example: Ethanol in water (alcohol + water mixture)
6) Solid in Liquid
👉 Example: Sugar in water, NaCl in water (saline water, concentration often written as g/L or mol/L)
👉 Example: Sugar in water, NaCl in water (saline water, concentration often written as g/L or mol/L)
C) Solid Solutions (Solvent = Solid)
7) Gas in Solid
👉 Example: Hydrogen in palladium (H2 absorbed in Pd)
👉 Example: Hydrogen in palladium (H2 absorbed in Pd)
8) Liquid in Solid
👉 Example: Amalgams like Na–Hg (sodium amalgam)
👉 Example: Amalgams like Na–Hg (sodium amalgam)
9) Solid in Solid
👉 Example: Alloys such as gold ornaments (Au mixed with Cu/Ag), brass (Cu + Zn)
👉 Example: Alloys such as gold ornaments (Au mixed with Cu/Ag), brass (Cu + Zn)
Did You Know?
✅ True solutions do not show
Tyndall effect
(they don’t scatter light), so a beam of light is not visible through them.✅ The concentration of a solution is commonly expressed as mol/L (molarity) , g/L , or % (w/v, v/v) depending on the need.
Q.2: Give an example of a solid solution in which the solute is a gas.
Answer
✅ 👉
Hydrogen gas dissolved in palladium
is a solid solution where the
solute = gas (H2)
and the
solvent = solid (Pd)
.
Explanation (Step by Step)
📌 Some metals (like
palladium
)
can absorb gases into their
crystal lattice
.🤔 So, H2 (gas) gets uniformly distributed inside Pd (solid) , forming a gas-in-solid solution .
Did You Know?
✅ Palladium can absorb a
surprisingly large amount of hydrogen
,
which is why it’s studied in
hydrogen storage research
.
Q.3: Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Answer
(i) Mole fraction (χ)
📌 Mole fraction is the ratio of moles of a component to the total moles of all components in the solution.
👉 For a binary solution (A + B):
• χA = nA / (nA + nB)
• χB = nB / (nA + nB)
✅ χA + χB = 1
🤔 Unit: No unit (dimensionless)
📌 Mole fraction is the ratio of moles of a component to the total moles of all components in the solution.
👉 For a binary solution (A + B):
• χA = nA / (nA + nB)
• χB = nB / (nA + nB)
✅ χA + χB = 1
🤔 Unit: No unit (dimensionless)
(ii) Molality (m)
📌 Molality is the number of moles of solute present in 1 kg of solvent .
👉 Formula:
• m = (Moles of solute) / (Mass of solvent in kg)
• m = nsolute / wsolvent(kg)
✅ Unit: mol kg−1 (or mol/kg)
📌 Molality is the number of moles of solute present in 1 kg of solvent .
👉 Formula:
• m = (Moles of solute) / (Mass of solvent in kg)
• m = nsolute / wsolvent(kg)
✅ Unit: mol kg−1 (or mol/kg)
(iii) Molarity (M)
📌 Molarity is the number of moles of solute present in 1 litre (1 L) of solution .
👉 Formula:
• M = (Moles of solute) / (Volume of solution in L)
• M = nsolute / Vsolution(L)
✅ Unit: mol L−1 (or mol/L)
📌 Molarity is the number of moles of solute present in 1 litre (1 L) of solution .
👉 Formula:
• M = (Moles of solute) / (Volume of solution in L)
• M = nsolute / Vsolution(L)
✅ Unit: mol L−1 (or mol/L)
(iv) Mass percentage (w/w %)
📌 Mass percentage of a component is the mass of that component in the solution divided by the total mass of solution, multiplied by 100 .
👉 For solute A in solvent B:
• Mass % (A) = [ wA / (wA + wB) ] × 100
✅ Unit: % (percentage)
📌 Mass percentage of a component is the mass of that component in the solution divided by the total mass of solution, multiplied by 100 .
👉 For solute A in solvent B:
• Mass % (A) = [ wA / (wA + wB) ] × 100
✅ Unit: % (percentage)
Explanation (Step by Step)
🤔 Many students confuse these terms, so remember this simple idea:📌 Mole fraction → compares moles with total moles (no unit)
📌 Molality → moles per kg of solvent (mol/kg), temperature independent
📌 Molarity → moles per L of solution (mol/L), changes with temperature (because volume changes)
📌 Mass % → compares mass with total mass (in %)
👉 Extra useful form (when masses are given):
• n = w(g) / M(g mol−1)
So molality can also be written as:
• m = wsolute(g) / [ Msolute(g mol−1) × wsolvent(kg) ]
Did You Know?
✅
Molality (m)
is preferred in many lab calculations because it does not change with temperature, while
molarity (M)
can change when the solution expands or contracts.✅ Mole fraction is super useful in topics like Raoult’s law and vapour pressure of solutions because it directly connects with gaseous behavior.
Q.4: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution.
What should be the molarity of such a sample if the density of the solution is 1.504 g mL−1?
Answer
✅ 👉 Molarity of
68% (w/w) HNO3 solution
=
16.23 mol L−1
(≈
16.2 M
)
Explanation (Step by Step)
📌 Given
• 68% (w/w) HNO3 means 68 g HNO3 in 100 g solution
• Density (ρ) = 1.504 g mL−1
• Molar mass of HNO3 = (1) + (14) + (3 × 16) = 63 g mol−1
• 68% (w/w) HNO3 means 68 g HNO3 in 100 g solution
• Density (ρ) = 1.504 g mL−1
• Molar mass of HNO3 = (1) + (14) + (3 × 16) = 63 g mol−1
Step 1: Find moles of HNO3 🤔
👉 n = w / M
n = 68 g / 63 g mol−1
n = 1.079 mol (approx)
👉 n = w / M
n = 68 g / 63 g mol−1
n = 1.079 mol (approx)
Step 2: Find volume of 100 g solution (using density) 📌
👉 Volume = Mass / Density
V = 100 g / 1.504 g mL−1
V = 66.5 mL
Convert to litres:
66.5 mL = 0.0665 L
👉 Volume = Mass / Density
V = 100 g / 1.504 g mL−1
V = 66.5 mL
Convert to litres:
66.5 mL = 0.0665 L
Step 3: Calculate molarity (M) ✅
👉 M = n / V(L)
M = 1.079 mol / 0.0665 L
M = 16.23 mol L−1
✅ Therefore, Molarity = 16.23 mol L−1 (≈ 16.2 M )
👉 M = n / V(L)
M = 1.079 mol / 0.0665 L
M = 16.23 mol L−1
✅ Therefore, Molarity = 16.23 mol L−1 (≈ 16.2 M )
Did You Know?
✅ “Concentrated” acids in labs are often given in
% (w/w)
,
not molarity.👉 To convert % (w/w) → molarity , density is essential because volume changes with concentration.
📌 That’s why the same acid can have different molarity if its density (or temperature) changes.
Q.5: A solution of glucose in water is labelled as 10% w/w. What would be the molality and mole fraction of each component in the solution?
If the density of solution is 1.2 g mL−1, then what shall be the molarity of the solution?
Answer
✅ 👉 For
10% (w/w) glucose solution
:• Molality (m) = 0.61 mol kg−1
• Mole fraction of glucose (χglucose) ≈ 0.01
• Mole fraction of water (χwater) ≈ 0.99
• Molarity (M) (given density 1.2 g mL−1 ) = 0.666 mol L−1 (≈ 0.67 M )
Explanation (Step by Step)
📌 Meaning of
10% w/w
:
10% (w/w) glucose ⇒ 10 g glucose present in 100 g solution.
So water = 100 − 10 = 90 g
10% (w/w) glucose ⇒ 10 g glucose present in 100 g solution.
So water = 100 − 10 = 90 g
Given 📌
• Glucose = C6H12O6 ,
• molar mass = (6×12) + (12×1) + (6×16) = 180 g mol−1
• Water = H2O , molar mass = (2×1) + 16 = 18 g mol−1
• Density of solution = 1.2 g mL−1
• Glucose = C6H12O6 ,
• molar mass = (6×12) + (12×1) + (6×16) = 180 g mol−1
• Water = H2O , molar mass = (2×1) + 16 = 18 g mol−1
• Density of solution = 1.2 g mL−1
(i) Molality (m) 🤔
👉 Molality = (moles of solute) / (mass of solvent in kg)
• Moles of glucose = 10 g / 180 g mol−1 = 0.0556 mol
• Mass of water = 90 g = 0.090 kg
So,
m = 0.0556 mol / 0.090 kg = 0.617 mol kg−1
✅ m ≈ 0.61 mol kg−1
👉 Molality = (moles of solute) / (mass of solvent in kg)
• Moles of glucose = 10 g / 180 g mol−1 = 0.0556 mol
• Mass of water = 90 g = 0.090 kg
So,
m = 0.0556 mol / 0.090 kg = 0.617 mol kg−1
✅ m ≈ 0.61 mol kg−1
(ii) Mole fraction of each component 📌
👉 First find moles:
• nglucose = 10/180 = 0.0556 mol
• nwater = 90/18 = 5.00 mol
Total moles = 0.0556 + 5.00 = 5.0556 mol
✅ Mole fraction of glucose:
χglucose = 0.0556 / 5.0556 = 0.011 ≈ 0.01
✅ Mole fraction of water:
χwater = 5.00 / 5.0556 = 0.989 ≈ 0.99
(Or simply: χwater = 1 − χglucose )
👉 First find moles:
• nglucose = 10/180 = 0.0556 mol
• nwater = 90/18 = 5.00 mol
Total moles = 0.0556 + 5.00 = 5.0556 mol
✅ Mole fraction of glucose:
χglucose = 0.0556 / 5.0556 = 0.011 ≈ 0.01
✅ Mole fraction of water:
χwater = 5.00 / 5.0556 = 0.989 ≈ 0.99
(Or simply: χwater = 1 − χglucose )
(iii) Molarity (M) using density ✅
🤔 We need volume of solution for 100 g solution .
👉 Volume = Mass / Density
V = 100 g / 1.2 g mL−1 = 83.33 mL = 0.08333 L
Now,
M = moles of glucose / volume of solution (L)
M = 0.0556 mol / 0.08333 L = 0.667 mol L−1
✅ M ≈ 0.666 mol L−1 (≈ 0.67 M )
🤔 We need volume of solution for 100 g solution .
👉 Volume = Mass / Density
V = 100 g / 1.2 g mL−1 = 83.33 mL = 0.08333 L
Now,
M = moles of glucose / volume of solution (L)
M = 0.0556 mol / 0.08333 L = 0.667 mol L−1
✅ M ≈ 0.666 mol L−1 (≈ 0.67 M )
Did You Know?
✅ Molality depends on
mass of solvent
,
so it stays almost constant even if temperature changes.✅ Molarity depends on volume of solution , so it can change with temperature (volume expands/contracts).
📌 That’s why density helps a lot when converting % w/w → molarity .